In the previous section, we saw the location of 'C' of some common objects. In this section we will see systems where particles are distributed in space
1. Consider a particle 'P' situated in space
• Let it be situated at a certain distance from the origin 'O' of the frame of reference
♦ x-axis is shown in red
♦ y-axis is shown in green
♦ z-axis is shown in blue
• This is shown in fig.7.42 (a) below:
2. The particle is shown as a yellow sphere
• It's mass is 0.5 kg
• We see that:
♦ It is not situated on the x-axis
♦ It is not situated on the y-axis
♦ It is not situated on the xy-plane
♦ It is situated in space
3. We know the method to specify the positions of such particles
• The method can be written in 5 steps:
(i) Drop a perpendicular from 'P' onto the xy-plane
• This is indicated by the blue line in fig.7.42(b) above
• Note that, the yellow sphere is a bit transparent. We see the blue line starting from the exact center of the sphere
(ii) Let P' be the foot of the perpendicular
• From P', draw a perpendicular onto the x-axis
• This is indicated by the green line
(iii) From P', draw a perpendicular onto the y-axis
• This is indicated by the red line
(iv) Let the length of the red line be xP = 2 m
• Let the length of the green line be yP = 3 m
• Let the length of the blue line be zP = 2 m
(v) Then the coordinates of P are (xP, yP, zP) = (2,3,2)
4. Consider another particle Q. It's mass is 0.75 kg. This is shown in fig.7.43(a) below:
• Q will have it's own red, green and blue lines
• Let the lengths of those lines be 1 m, 2 m, and 1 m respectively
• So the coordinates (xQ, yQ, zQ) will be (1,2,1)
5. So we have a system consisting of two particles P and Q
• We want the location of the 'C' of this system
6. In such cases, the 'C' lies at a point whose coordinates are (X,Y,Z)
• We already know the method to find 'X' and 'Y'
7. We can apply the same method to find 'Z'
• But to find 'Z', the method should be applied in the z-direction
• As usual, the method involves only 4 steps:
(i) Take the distances (from the xy-plane) for each particle
(ii) Apply the 'due weightage'
(iii) Find the average of those 'weighted distances'
(iv) This average is the 'Z'
8. So we can write a new equation:
Eq.7.3: $\mathbf\small{Z=\frac{\sum{m_iz_i} }{\sum{m_i}}}$
9. Now we can form the table:
10. From the table, we get:
$\mathbf\small{\sum{m_i}=1.25}$
$\mathbf\small{\sum{m_ix_i}=1.75}$
$\mathbf\small{\sum{m_iy_i}=3}$
$\mathbf\small{\sum{m_iz_i}=1.75}$
• Thus we get:
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{1.75}{1.25}=1.4\, \text{m}}$
$\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{3}{1.25}=2.4\, \text{m}}$
$\mathbf\small{Z=\frac{\sum{m_iz_i} }{\sum{m_i}}=\frac{1.75}{1.25}=1.4\, \text{m}}$
11. Using these coordinates, we can mark 'C'. This is shown as a small white sphere in fig.7.43(b) above
• We see that, 'C' has it's own red, green and blue lines
♦ The length of it's red line = X = 1.4 m
♦ The length of it's green line = Y = 2.4 m
♦ The length of it's blue line = Z = 1.4 m
• Also note that, the 'C' lies on the line joining P and Q
• In such 3-dimensional problems, another easier method can be used. Let us see the details of that method:
1. In fig.7.44(a) below, a vector is shown in magenta color
• It's tail end coincides with 'O'
• It's tip coincides with the center of 'P'
■ So it is the position vector of P
• We can denote it as $\mathbf\small{\vec{r}_P}$. (Details here)
2. Imagine that, a person wants to go from O to P
• He can take either one of the two paths given below:
Path 1:
• This path is along the $\mathbf\small{\vec{r}_P}$. It starts from O and ends at P
Path 2:
This path has 3 segments:
■ Segment 1:
• This segment starts from O
• The person travels a distance xP
♦ xP is the length of the red line
♦ So 'magnitude' of travel is xP
• Direction of travel for this segment is: 'along the x-axis'
• So the magnitude and direction of this travel can be represented in vector form as: $\mathbf\small{(x_P)\hat{i}}$
• At the end of this travel, the person is at the foot of the green line
■ Segment 2:
• This segment starts from the foot of the green line
• The person travels a distance yP
♦ yP is the length of the green line
♦ So 'magnitude' of travel is yP
• Direction of travel for this segment is: 'parallel to the y-axis'
• So the magnitude and direction of this travel can be represented in vector form as: $\mathbf\small{(y_P)\hat{j}}$
• At the end of this travel, the person is at the foot of the blue line
■ Segment 3:
• This segment starts from the foot of the blue line
• The person travels a distance zP
♦ zP is the length of the blue line
♦ So 'magnitude' of travel is zP
• Direction of travel for this segment is: 'parallel to the z-axis'
• So the magnitude and direction of this travel can be represented in vector form as: $\mathbf\small{(z_P)\hat{k}}$
• At the end of this travel, the person reaches the particle P
3. So we see 3 vectors in path 2. If we add those vectors, we will reach P
• That means:
Path 2 = $\mathbf\small{(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}}$
4. The initial and final points in both paths 1 and 2 are the same
• So we can write: Path 1 = Path 2
• That is: $\mathbf\small{\vec{r}_P=(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}}$
5. The above equality in (4) can be established based on fig.7.44(b) also:
(i) In fig.7.44(b), there are a total of 5 vectors:
Red, green, blue, yellow and magenta
(ii) We want to prove this:
Magenta = red + green + blue
(iii) Consider the following three vectors:
Red, green and yellow
• Applying triangle law of vector addition, we get:
yellow = red + green
(iv) Consider the following three vectors:
Yellow, blue and magenta
• Applying triangle law of vector addition, we get:
Magenta = yellow + blue
(v) Now we expand 'yellow' using the result in (iii). We get:
Magenta = red + green + blue
• This is the required result that we mentioned in (ii)
6. So it is proved beyond doubt. We can confidently write:
$\mathbf\small{\vec{r}_P=(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}}$
• In this equation,
♦ $\mathbf\small{\vec{r}_P}$ is the position vector
♦ xP, yP and zP are the coordinates of P
■ So we can write:
The following two items are closely related:
(i) position vector of a particle
(ii) The coordinates of that particle
1. Consider a particle 'P' situated in space
• Let it be situated at a certain distance from the origin 'O' of the frame of reference
♦ x-axis is shown in red
♦ y-axis is shown in green
♦ z-axis is shown in blue
• This is shown in fig.7.42 (a) below:
 |
| Fig.7.42 |
• It's mass is 0.5 kg
• We see that:
♦ It is not situated on the x-axis
♦ It is not situated on the y-axis
♦ It is not situated on the xy-plane
♦ It is situated in space
3. We know the method to specify the positions of such particles
• The method can be written in 5 steps:
(i) Drop a perpendicular from 'P' onto the xy-plane
• This is indicated by the blue line in fig.7.42(b) above
• Note that, the yellow sphere is a bit transparent. We see the blue line starting from the exact center of the sphere
(ii) Let P' be the foot of the perpendicular
• From P', draw a perpendicular onto the x-axis
• This is indicated by the green line
(iii) From P', draw a perpendicular onto the y-axis
• This is indicated by the red line
(iv) Let the length of the red line be xP = 2 m
• Let the length of the green line be yP = 3 m
• Let the length of the blue line be zP = 2 m
(v) Then the coordinates of P are (xP, yP, zP) = (2,3,2)
4. Consider another particle Q. It's mass is 0.75 kg. This is shown in fig.7.43(a) below:
 |
| Fig.7.43 |
• Let the lengths of those lines be 1 m, 2 m, and 1 m respectively
• So the coordinates (xQ, yQ, zQ) will be (1,2,1)
5. So we have a system consisting of two particles P and Q
• We want the location of the 'C' of this system
6. In such cases, the 'C' lies at a point whose coordinates are (X,Y,Z)
• We already know the method to find 'X' and 'Y'
7. We can apply the same method to find 'Z'
• But to find 'Z', the method should be applied in the z-direction
• As usual, the method involves only 4 steps:
(i) Take the distances (from the xy-plane) for each particle
(ii) Apply the 'due weightage'
(iii) Find the average of those 'weighted distances'
(iv) This average is the 'Z'
8. So we can write a new equation:
Eq.7.3: $\mathbf\small{Z=\frac{\sum{m_iz_i} }{\sum{m_i}}}$
9. Now we can form the table:
$\mathbf\small{\sum{m_i}=1.25}$
$\mathbf\small{\sum{m_ix_i}=1.75}$
$\mathbf\small{\sum{m_iy_i}=3}$
$\mathbf\small{\sum{m_iz_i}=1.75}$
• Thus we get:
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{1.75}{1.25}=1.4\, \text{m}}$
$\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{3}{1.25}=2.4\, \text{m}}$
$\mathbf\small{Z=\frac{\sum{m_iz_i} }{\sum{m_i}}=\frac{1.75}{1.25}=1.4\, \text{m}}$
11. Using these coordinates, we can mark 'C'. This is shown as a small white sphere in fig.7.43(b) above
• We see that, 'C' has it's own red, green and blue lines
♦ The length of it's red line = X = 1.4 m
♦ The length of it's green line = Y = 2.4 m
♦ The length of it's blue line = Z = 1.4 m
• Also note that, the 'C' lies on the line joining P and Q
• Now we know the method to find 'C' of the particles distributed in 3 dimensional space
1. In fig.7.44(a) below, a vector is shown in magenta color
 |
| Fig.7.44 |
• It's tip coincides with the center of 'P'
■ So it is the position vector of P
• We can denote it as $\mathbf\small{\vec{r}_P}$. (Details here)
2. Imagine that, a person wants to go from O to P
• He can take either one of the two paths given below:
Path 1:
• This path is along the $\mathbf\small{\vec{r}_P}$. It starts from O and ends at P
Path 2:
This path has 3 segments:
■ Segment 1:
• This segment starts from O
• The person travels a distance xP
♦ xP is the length of the red line
♦ So 'magnitude' of travel is xP
• Direction of travel for this segment is: 'along the x-axis'
• So the magnitude and direction of this travel can be represented in vector form as: $\mathbf\small{(x_P)\hat{i}}$
• At the end of this travel, the person is at the foot of the green line
■ Segment 2:
• This segment starts from the foot of the green line
• The person travels a distance yP
♦ yP is the length of the green line
♦ So 'magnitude' of travel is yP
• Direction of travel for this segment is: 'parallel to the y-axis'
• So the magnitude and direction of this travel can be represented in vector form as: $\mathbf\small{(y_P)\hat{j}}$
• At the end of this travel, the person is at the foot of the blue line
■ Segment 3:
• This segment starts from the foot of the blue line
• The person travels a distance zP
♦ zP is the length of the blue line
♦ So 'magnitude' of travel is zP
• Direction of travel for this segment is: 'parallel to the z-axis'
• So the magnitude and direction of this travel can be represented in vector form as: $\mathbf\small{(z_P)\hat{k}}$
• At the end of this travel, the person reaches the particle P
3. So we see 3 vectors in path 2. If we add those vectors, we will reach P
• That means:
Path 2 = $\mathbf\small{(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}}$
4. The initial and final points in both paths 1 and 2 are the same
• So we can write: Path 1 = Path 2
• That is: $\mathbf\small{\vec{r}_P=(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}}$
5. The above equality in (4) can be established based on fig.7.44(b) also:
(i) In fig.7.44(b), there are a total of 5 vectors:
Red, green, blue, yellow and magenta
(ii) We want to prove this:
Magenta = red + green + blue
(iii) Consider the following three vectors:
Red, green and yellow
• Applying triangle law of vector addition, we get:
yellow = red + green
(iv) Consider the following three vectors:
Yellow, blue and magenta
• Applying triangle law of vector addition, we get:
Magenta = yellow + blue
(v) Now we expand 'yellow' using the result in (iii). We get:
Magenta = red + green + blue
• This is the required result that we mentioned in (ii)
6. So it is proved beyond doubt. We can confidently write:
$\mathbf\small{\vec{r}_P=(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}}$
• In this equation,
♦ $\mathbf\small{\vec{r}_P}$ is the position vector
♦ xP, yP and zP are the coordinates of P
■ So we can write:
The following two items are closely related:
(i) position vector of a particle
(ii) The coordinates of that particle
7. Now, if 'Q' is another particle with coordinates (xQ,yQ,zQ), we can write:
$\mathbf\small{\vec{r}_Q=(x_Q)\hat{i}+(y_Q)\hat{j}+(z_Q)\hat{k}}$
8. Consider the system consisting of the two particles P and Q
• Let 'C' be the center of mass of the system
• The coordinates of 'C' are (X,Y,Z)
9. If we know those coordinates, we can easily write the position vector of 'C'
• Based on (6) above, we get: $\mathbf\small{\vec{r}_C=(X)\hat{i}+(Y)\hat{j}+(Z)\hat{k}}$
10. Now we expand X, Y and Z. We have:
• $\mathbf\small{X=\frac{m_p\,x_P+m_Q\,x_Q}{m_P+m_Q}=\frac{m_p\,x_P+m_Q\,x_Q}{M}}$
• $\mathbf\small{Y=\frac{m_p\,y_P+m_Q\,y_Q}{M}}$
• $\mathbf\small{Z=\frac{m_p\,z_P+m_Q\,z_Q}{M}}$
11. So the equation in (9) becomes:
$\mathbf\small{\vec{r}_C=\left[\frac{m_p\,x_P+m_Q\,x_Q}{M} \right]\hat{i}+\left[\frac{m_p\,y_P+m_Q\,y_Q}{M} \right]\hat{j}+\left[\frac{m_p\,z_P+m_Q\,z_Q}{M} \right]\hat{k}}$
$\mathbf\small{\Rightarrow \vec{r}_C=\frac{(m_P\,x_P)\hat{i}+(m_Q\,x_Q)\hat{i}+(m_P\,y_P)\hat{j}+(m_Q\,y_Q)\hat{j}+(m_P\,z_P)\hat{k}+(m_Q\,z_Q)\hat{k}}{M}}$
12. Now we rearrange the above equation by the two steps:
(i) Bringing the terms with mP together
(ii) Bringing the terms with mQ together
• We get:
$\mathbf\small{\vec{r}_C=\frac{[(m_P\,x_P)\hat{i}+(m_P\,y_P)\hat{j}+(m_P\,z_P)\hat{k}]+[(m_Q\,x_Q)\hat{i}+(m_Q\,y_Q)\hat{j}+(m_Q\,z_Q)\hat{k}]}{M}}$
• Taking mP and mQ outside, we get:
$\mathbf\small{\vec{r}_C=\frac{m_P[(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}]+m_Q[(x_Q)\hat{i}+(y_Q)\hat{j}+(z_Q)\hat{k}]}{M}}$
13. But:
♦ $\mathbf\small{(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}=\vec{r}_P}$
♦ $\mathbf\small{(x_Q)\hat{i}+(y_Q)\hat{j}+(z_Q)\hat{k}=\vec{r}_Q}$
• Thus (12) becomes:
$\mathbf\small{\vec{r}_C=\frac{m_P[\vec{r}_P]+m_Q[\vec{r}_Q]}{M}}$
14. This gives us an easy method to write the 'position vector of C'
• Let us check and see if we will get the same answer as before:
(i) The coordinates of P are (2,3,2)
• So $\mathbf\small{\vec{r}_P=2\hat{i}+3\hat{j}+2\hat{k}}$
• So $\mathbf\small{m_P\,\vec{r}_P=0.5(2\hat{i}+3\hat{j}+2\hat{k})=\hat{i}+1.5\hat{j}+\hat{k}}$
(ii) The coordinates of Q are (1,2,1)
• So $\mathbf\small{\vec{r}_Q=\hat{i}+2\hat{j}+\hat{k}}$
• So $\mathbf\small{m_Q\,\vec{r}_P=0.75(\hat{i}+2\hat{j}+\hat{k})=0.75\hat{i}+1.5\hat{j}+0.75\hat{k}}$
(iii) Thus the numerator in (13) becomes:
$\mathbf\small{(\hat{i}+1.5\hat{j}+\hat{k})+(0.75\hat{i}+1.5\hat{j}+0.75\hat{k})}$
$\mathbf\small{=(1.75\hat{i}+3.0\hat{j}+1.75\hat{k})}$
• The denominator M is the total mass = (mP + mQ) = (0.5+0.75) = 1.25 kg
(iv) Thus we get:
$\mathbf\small{\vec{r}_C=\frac{1.75\hat{i}+3.0\hat{j}+1.75\hat{k}}{1.25}=1.4\hat{i}+2.4\hat{j}+1.4\hat{k}}$
(v) So the coordinates of C are: (1.4,2.4,1.4)
• This is the same result that we obtained before
• P, Q, R, S, . . . are various particles in a system
• There are a total of 'n' particles in the system
♦ P is the first particle
♦ Q is the second particle
♦ R is the third particle . . . so on . . .
• Then:
♦ mP is the first mass
♦ mQ is the second mass
♦ mR is the third mass . . . so on . . .
16. We can write:
♦ m1 which is equal to mP is the first mass
♦ m2 which is equal to mQ is the second mass
♦ m3 which is equal to mR is the third mass . . . so on . . .
• Then:
♦ mi is the ith mass
♦ mn is the last mass
17. Also we can write:
♦ $\mathbf\small{\vec{r}_1}$ which is equal to $\mathbf\small{\vec{r}_P}$ is the first position vector
♦ $\mathbf\small{\vec{r}_2}$ which is equal to $\mathbf\small{\vec{r}_Q}$ is the second position vector
♦ $\mathbf\small{\vec{r}_3}$ which is equal to $\mathbf\small{\vec{r}_R}$ is the third position vector . . . so on . . .
• Then:
♦ $\mathbf\small{\vec{r}_i}$ is the ith position vector
♦ $\mathbf\small{\vec{r}_n}$ is the last position vector
18. So the equation in (13) becomes:
$\mathbf\small{\vec{r}_C=\frac{m_1\,\vec{r}_1+m_2\,\vec{r}_2\,+\,.\,.\,.\,+\,m_i\,\vec{r}_i+\,.\,.\,.\,+\,m_n\,\vec{r}_n}{m_1+m_2\,+\,.\,.\,.\,+\,m_i+\,.\,.\,.\,+\,m_n}}$
$\mathbf\small{\Rightarrow \vec{r}_C=\frac{m_1\,\vec{r}_1+m_2\,\vec{r}_2\,+\,.\,.\,.\,+\,m_i\,\vec{r}_i+\,.\,.\,.\,+\,m_n\,\vec{r}_n}{M}}$
19. Thus we get
Eq.7.4: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$
• This equation is applicable to 2- dimensional and 1-dimensional problems as well
• We will consider solved example 7.3. We will do it as a new solved example:
Solved example 7.7
Three particles P, Q and R are situated at the vertices of an equilateral triangle of side 0.5 m. Their masses are 100 g, 150 g and 200 g respectively. Find the location of the 'C' of this system of 3 particles
Solution:
1. We have three particles so situated in space that, they are at the vertices of an equilateral triangle
• For ease in calculations, we arrange the frame of reference in the following way:
♦ The plane of the triangle lies in the xy-plane
♦ One of the sides (say PQ), lies on the x-axis
♦ The left vertex (P) of that side coincides with O
• This arrangement is shown in fig.7.35 below:
2. Once this arrangement is fixed, we can easily write the coordinates of the bottom vertices
• The coordinates of P will be (0,0)
• The coordinates of Q will be (0.5,0)
• The x-coordinate of R will be 0.25
3. To find the y-coordinate of R, we use the following steps:
• Drop a perpendicular from R. This is shown in fig.b
• The foot of this perpendicular is R'
• In the right triangle PR'R, we have:
$\mathbf\small{\sin 60=\frac{RR'}{PR}=\frac{RR'}{0.5}}$
$\mathbf\small{\Rightarrow RR'=\sin 60 \times 0.5=0.433}$
So the y-coordinate of R is 0.433
4. Once we obtain the coordinates, we can write the position vectors:
(i) $\mathbf\small{\vec{r}_P=0\hat{i}+0\hat{j}}$
• This is a null vector
(ii) $\mathbf\small{\vec{r}_Q=0.5\hat{i}+0\hat{j}}$
(iii) $\mathbf\small{\vec{r}_R=0.25\hat{i}+0.433\hat{j}}$
5. We have: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$
• The numerator becomes: $\mathbf\small{0.1 \times 0+0.15 \times 0.5\hat{i}+0.2 \times 0.25\hat{i}+0.2 \times 0.433\hat{j}}$
= $\mathbf\small{0.125\hat{i}+0.0866\hat{j}}$
• The denominator is the total mass M = 0.450 kg
6. Thus we get: $\mathbf\small{\vec{r}_C=\frac{0.125\hat{i}+0.0866\hat{j}}{0.45}=0.278 \hat{i}+0.192 \hat{j}}$
7. So the coordinates of 'C' are: (0.278,0.192)
• This is the same result that we obtained earlier
• Note that, the unit vector $\mathbf\small{\hat{k}}$ does not come in the calculations because, this is a 2-dimensional problem
$\mathbf\small{\vec{r}_Q=(x_Q)\hat{i}+(y_Q)\hat{j}+(z_Q)\hat{k}}$
8. Consider the system consisting of the two particles P and Q
• Let 'C' be the center of mass of the system
• The coordinates of 'C' are (X,Y,Z)
9. If we know those coordinates, we can easily write the position vector of 'C'
• Based on (6) above, we get: $\mathbf\small{\vec{r}_C=(X)\hat{i}+(Y)\hat{j}+(Z)\hat{k}}$
10. Now we expand X, Y and Z. We have:
• $\mathbf\small{X=\frac{m_p\,x_P+m_Q\,x_Q}{m_P+m_Q}=\frac{m_p\,x_P+m_Q\,x_Q}{M}}$
• $\mathbf\small{Y=\frac{m_p\,y_P+m_Q\,y_Q}{M}}$
• $\mathbf\small{Z=\frac{m_p\,z_P+m_Q\,z_Q}{M}}$
11. So the equation in (9) becomes:
$\mathbf\small{\vec{r}_C=\left[\frac{m_p\,x_P+m_Q\,x_Q}{M} \right]\hat{i}+\left[\frac{m_p\,y_P+m_Q\,y_Q}{M} \right]\hat{j}+\left[\frac{m_p\,z_P+m_Q\,z_Q}{M} \right]\hat{k}}$
$\mathbf\small{\Rightarrow \vec{r}_C=\frac{(m_P\,x_P)\hat{i}+(m_Q\,x_Q)\hat{i}+(m_P\,y_P)\hat{j}+(m_Q\,y_Q)\hat{j}+(m_P\,z_P)\hat{k}+(m_Q\,z_Q)\hat{k}}{M}}$
12. Now we rearrange the above equation by the two steps:
(i) Bringing the terms with mP together
(ii) Bringing the terms with mQ together
• We get:
$\mathbf\small{\vec{r}_C=\frac{[(m_P\,x_P)\hat{i}+(m_P\,y_P)\hat{j}+(m_P\,z_P)\hat{k}]+[(m_Q\,x_Q)\hat{i}+(m_Q\,y_Q)\hat{j}+(m_Q\,z_Q)\hat{k}]}{M}}$
• Taking mP and mQ outside, we get:
$\mathbf\small{\vec{r}_C=\frac{m_P[(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}]+m_Q[(x_Q)\hat{i}+(y_Q)\hat{j}+(z_Q)\hat{k}]}{M}}$
13. But:
♦ $\mathbf\small{(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}=\vec{r}_P}$
♦ $\mathbf\small{(x_Q)\hat{i}+(y_Q)\hat{j}+(z_Q)\hat{k}=\vec{r}_Q}$
• Thus (12) becomes:
$\mathbf\small{\vec{r}_C=\frac{m_P[\vec{r}_P]+m_Q[\vec{r}_Q]}{M}}$
14. This gives us an easy method to write the 'position vector of C'
• Let us check and see if we will get the same answer as before:
(i) The coordinates of P are (2,3,2)
• So $\mathbf\small{\vec{r}_P=2\hat{i}+3\hat{j}+2\hat{k}}$
• So $\mathbf\small{m_P\,\vec{r}_P=0.5(2\hat{i}+3\hat{j}+2\hat{k})=\hat{i}+1.5\hat{j}+\hat{k}}$
(ii) The coordinates of Q are (1,2,1)
• So $\mathbf\small{\vec{r}_Q=\hat{i}+2\hat{j}+\hat{k}}$
• So $\mathbf\small{m_Q\,\vec{r}_P=0.75(\hat{i}+2\hat{j}+\hat{k})=0.75\hat{i}+1.5\hat{j}+0.75\hat{k}}$
(iii) Thus the numerator in (13) becomes:
$\mathbf\small{(\hat{i}+1.5\hat{j}+\hat{k})+(0.75\hat{i}+1.5\hat{j}+0.75\hat{k})}$
$\mathbf\small{=(1.75\hat{i}+3.0\hat{j}+1.75\hat{k})}$
• The denominator M is the total mass = (mP + mQ) = (0.5+0.75) = 1.25 kg
(iv) Thus we get:
$\mathbf\small{\vec{r}_C=\frac{1.75\hat{i}+3.0\hat{j}+1.75\hat{k}}{1.25}=1.4\hat{i}+2.4\hat{j}+1.4\hat{k}}$
(v) So the coordinates of C are: (1.4,2.4,1.4)
• This is the same result that we obtained before
15. We will generalize this method:
• There are a total of 'n' particles in the system
♦ P is the first particle
♦ Q is the second particle
♦ R is the third particle . . . so on . . .
• Then:
♦ mP is the first mass
♦ mQ is the second mass
♦ mR is the third mass . . . so on . . .
16. We can write:
♦ m1 which is equal to mP is the first mass
♦ m2 which is equal to mQ is the second mass
♦ m3 which is equal to mR is the third mass . . . so on . . .
• Then:
♦ mi is the ith mass
♦ mn is the last mass
17. Also we can write:
♦ $\mathbf\small{\vec{r}_1}$ which is equal to $\mathbf\small{\vec{r}_P}$ is the first position vector
♦ $\mathbf\small{\vec{r}_2}$ which is equal to $\mathbf\small{\vec{r}_Q}$ is the second position vector
♦ $\mathbf\small{\vec{r}_3}$ which is equal to $\mathbf\small{\vec{r}_R}$ is the third position vector . . . so on . . .
• Then:
♦ $\mathbf\small{\vec{r}_i}$ is the ith position vector
♦ $\mathbf\small{\vec{r}_n}$ is the last position vector
18. So the equation in (13) becomes:
$\mathbf\small{\vec{r}_C=\frac{m_1\,\vec{r}_1+m_2\,\vec{r}_2\,+\,.\,.\,.\,+\,m_i\,\vec{r}_i+\,.\,.\,.\,+\,m_n\,\vec{r}_n}{m_1+m_2\,+\,.\,.\,.\,+\,m_i+\,.\,.\,.\,+\,m_n}}$
$\mathbf\small{\Rightarrow \vec{r}_C=\frac{m_1\,\vec{r}_1+m_2\,\vec{r}_2\,+\,.\,.\,.\,+\,m_i\,\vec{r}_i+\,.\,.\,.\,+\,m_n\,\vec{r}_n}{M}}$
19. Thus we get
Eq.7.4: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$
• This equation is applicable to 2- dimensional and 1-dimensional problems as well
• Let us apply it to a two dimensional problem that we solved in the previous section
Solved example 7.7
Three particles P, Q and R are situated at the vertices of an equilateral triangle of side 0.5 m. Their masses are 100 g, 150 g and 200 g respectively. Find the location of the 'C' of this system of 3 particles
Solution:
1. We have three particles so situated in space that, they are at the vertices of an equilateral triangle
• For ease in calculations, we arrange the frame of reference in the following way:
♦ The plane of the triangle lies in the xy-plane
♦ One of the sides (say PQ), lies on the x-axis
♦ The left vertex (P) of that side coincides with O
• This arrangement is shown in fig.7.35 below:
 |
| Fig.7.35 |
• The coordinates of P will be (0,0)
• The coordinates of Q will be (0.5,0)
• The x-coordinate of R will be 0.25
3. To find the y-coordinate of R, we use the following steps:
• Drop a perpendicular from R. This is shown in fig.b
• The foot of this perpendicular is R'
• In the right triangle PR'R, we have:
$\mathbf\small{\sin 60=\frac{RR'}{PR}=\frac{RR'}{0.5}}$
$\mathbf\small{\Rightarrow RR'=\sin 60 \times 0.5=0.433}$
So the y-coordinate of R is 0.433
4. Once we obtain the coordinates, we can write the position vectors:
(i) $\mathbf\small{\vec{r}_P=0\hat{i}+0\hat{j}}$
• This is a null vector
(ii) $\mathbf\small{\vec{r}_Q=0.5\hat{i}+0\hat{j}}$
(iii) $\mathbf\small{\vec{r}_R=0.25\hat{i}+0.433\hat{j}}$
5. We have: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$
• The numerator becomes: $\mathbf\small{0.1 \times 0+0.15 \times 0.5\hat{i}+0.2 \times 0.25\hat{i}+0.2 \times 0.433\hat{j}}$
= $\mathbf\small{0.125\hat{i}+0.0866\hat{j}}$
• The denominator is the total mass M = 0.450 kg
6. Thus we get: $\mathbf\small{\vec{r}_C=\frac{0.125\hat{i}+0.0866\hat{j}}{0.45}=0.278 \hat{i}+0.192 \hat{j}}$
7. So the coordinates of 'C' are: (0.278,0.192)
• This is the same result that we obtained earlier
• Note that, the unit vector $\mathbf\small{\hat{k}}$ does not come in the calculations because, this is a 2-dimensional problem
So now we know how to find the location of the 'C' of any given system. In the next section, we will see the significance of 'C'
No comments:
Post a Comment